Question 155

Two pipes can fill a tank in 10 hours and 16 hours respectively. A third pipe can empty the tank in 32 hours. If all the three pipes are opened simultaneously then in how much time the tank will be full ? (in hours)

Solution

Total quantity = 160 units
Speed pipe 1 = 16 $$\frac{units}{hours}$$

Speed pipe 2 = 10 $$\frac{units}{hours}$$

Speed pipe 3 = 5 $$\frac{units}{hours}$$

Time reqd = Total Units / [Speed (pipe 1 + pipe 2) - Speed Pipe 3]

Time = 160/21 = 7 $$\frac{13}{21}$$ hours

Create a FREE account and get: